Physics Test - 34

PE of simple harmonic oscilator
\(=\frac{1}{2} k x^{2}=\frac{1}{2} m \omega^{2} x^{2}\)
where, \(k=\) force constant
\(=\mathrm{m} \mathrm{w}^{2}\)
When PE is piotted against dispiacement \(x\), we will cotain a perabola
When \(x=0 \quad\) PE \(=0\)
When \(x=\pm A \quad P E=\) maximum
\(=\frac{1}{2} m \omega^{2} A^{2}\)
KB of a simple harmonie oseillator \(=\frac{1}{2} \mathrm{mv}^{2}\)
\((\because v=w \sqrt{A^{2}-x^{2}})\)
\(=\frac{1}{2} m[\omega \sqrt{A^{2}-x^{2}}]^{2}\)
\(=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)\)
This is also parabola, if plot KE against displacement \(x\)
10
\(\mathrm{KE}=0\) at \(\mathrm{x}=\pm \mathrm{A}\)
and \(\mathrm{KE}=\frac{1}{2} \mathrm{mod}^{2} \mathrm{A}^{2}\) at \(\mathrm{x}=0\)
Now, total energy of the simple harmonic oscillator \(=P E+K E\)

[using Eqs. (i) and (ii)]
\(=\frac{1}{2} \mathrm{m} \omega^{2} x^{2}+\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)\)
\(=\frac{1}{2} \mathrm{m} \omega^{2} x^{2}+\frac{1}{2} m \omega^{2} A^{2}-\frac{1}{2} m \omega^{2} x^{2}\)
\(\mathrm{TE}=\frac{1}{2} \mathrm{m} \omega^{2} A^{2}\)
which is constant and does not depend on \(x\)
Plotting under the above guidelines \(\mathrm{KE},\) PE and TE versus displacement x-graph as follows:

Given, magnetic moment of bar magnet \(\mathrm{M}=0.48 \mathrm{J} / \mathrm{T}\)
Distance from the centre of magnet \(\mathrm{d}=10 \mathrm{cm}=0.1 \mathrm{m}\)
Use the formula of magnetic field due to a short bar magnet on its equatorial line.
\(\therefore\) Magnetic field at point \(P\)

\[
\begin{aligned}
\mathbf{B} &=\frac{\mu_{0}}{4 \pi} \cdot \frac{\mathbf{M}}{\mathrm{d}^{3}} \\
&=10^{-7} \times \frac{0.48}{(0.1)^{3}}=0.48 \times 10^{-4} \mathrm{T}
\end{aligned}
\]
The direction of magnetic field on equitorial line is opposite to the direction of magnetic moment. So, the direction of magnetic field is from \(\mathrm{N}\) to \(\mathrm{S}\) pole of the magnet

In one hour, hour hand moves by 1 hour distance on clock while minute hand moves by 12 hour distance. marked on clock
thus \(\frac{w_{1}}{w_{2}}=\frac{2 \pi}{1} \times \frac{12}{2 \pi}\)
\(\Rightarrow \frac{w_{1}}{w_{2}}=\frac{12}{1}\)
KEY CONCEPTS
Angular displacement Let's say, a body is moving in circular direction as shown in fig.

For \(\theta\) angle and the radius of curved path is \(r\). The linear displacement is related to the angular displacement as: \(s=r \theta\)
\(\theta=\frac{s}{r}\)
(i)
Where \(r=\) radius of the curvature
\(\theta=\) angular displacement.
The equation (i) is the angular displacement equation.
Angular velocity The angular velocity is the measure of how fast a body is changing its angle w.r.t a point the angular velocity is measured in radians per second. The angular velocity is represented by \(\omega\) and hence \(\omega=\frac{\theta}{t}\)

White spot on screen would be central maximum where
\(\Delta \mathrm{x}=0 \quad \mathrm{y}=\frac{\mathrm{d}}{2}-\frac{\mathrm{d}}{8}=\frac{3 \mathrm{d}}{8}\)

at the lowest position
effective pull \(=m g+2 m g\)
\(=3 \mathrm{mg}\)
\(Y=\frac{3 m g l}{A \Delta l}\)
\(\Rightarrow \frac{\Delta l}{l}=\frac{3 m g}{A Y}\)
\(\frac{\Delta l}{l}=\frac{3 \times 9.8}{3 \times 10^{-6} \times 10^{11}}\)
\[
=0.9 \times 10^{-4}
\]

using conservation of energy
\[
\begin{array}{l}
-\frac{\mathrm{GMm}}{2 \mathrm{R}}+\frac{1}{2} \mathrm{mv}^{2}=0 \\
\text { or } \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{\mathrm{R}} \\
\text { or } \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}
\end{array}
\]
KEY CONCEPTS
Gravitational potential energy
(i) This is negative of work done by gravitational forces in making the system from infinite separation to the present position.
(ii) Gravitational potential energy of two point masses, is
\[
\mathrm{U}=-\frac{\mathrm{Gm}_{1} \mathrm{m}_{2}}{\mathrm{r}}
\]
(lii) To find gravitational potential energy of more than two point masses we have to make pair of masses. Neither of the pair should be repeated. For example, in case of four point masses,
For n point masses, total number of pairs will be \(\frac{n(n-1)}{2}\)
(iv) If a point mass \(\mathrm{m}\) is placed on the surface of earth, the potential energy here is
\[
\mathrm{U}_{0}=\frac{\mathrm{GMm}}{\mathrm{R}}
\]

and potential energy at height \(h\) is \(\mathrm{U}_{\mathrm{h}}=-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{b})}\)
The difference in potential energy would be
\[
\begin{array}{l}
\Delta \mathrm{U}=\mathrm{U}_{\mathrm{h}}-\mathrm{U}_{0} \text { or } \Delta \mathrm{U}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{s}}{\mathrm{p}}} \\
\text { if } \mathrm{h}<<\mathrm{R}, \Delta \mathrm{U} \approx \mathrm{mgh}
\end{array}
\]
Escape velocity
(i) From the surface of earth,
\[
\begin{aligned}
v_{e} &=\sqrt{2 g R} \\
&=\sqrt{\frac{2 G M}{R}}\left(\text { as } g=\frac{G M}{R^{2}}\right) \\
& \approx 11 \cdot 2 \mathrm{kms}^{-1}
\end{aligned}
\]
(ii) Escape velocity does not depend upon the angle at which particle is projected from the surface.
Kinetic Energy For a body of mass 'm' moving with velocity ' \(\mathrm{v}\) '. The kinetic energy is given by
\[
\mathrm{K} . \mathrm{E} .=\frac{1}{2} \mathrm{mv}^{2}
\]
Gravitational Potential The gravitational potential energy of a body at a point is defined as the amount of work done in bringing the body from infinity to that point against the gravitational force.

\(\mathbf{W}=\int_{\infty}^{\mathrm{r}} \frac{\mathrm{GMm}}{\mathrm{x}^{2}} \mathrm{d} \mathrm{x}=-\mathrm{GMm}\left[\frac{1}{\mathrm{x}}\right]_{\infty}^{\mathrm{r}}\)
\(\mathbf{W}=-\frac{\mathrm{GMm}}{\mathrm{r}}\)
This work done is stored inside the body as its gravitational potential energy
\(\therefore \mathrm{U}=-\frac{\mathrm{GMm}}{\mathrm{r}}\)
(i) Potential energy is a scalar quantity.
(ii) Unit : Joule
(iii) Dimension : \(\left[\mathrm{ML}^{2} \mathrm{T}^{-2}\right]\)
(iv) Gravitational potential energy is always negative in the gravitational field because the force is always attractive in nature.
(v) As the distance \(r\) increases, the gravitational potential energy becomes less negative ie., it increases.
(vi) If \(\mathbf{r}=\infty\) then it becomes zero (maximum)
(vii) in case of discrete distribution of masses
Gravitational potential energy
\(\mathbf{U}=\sum \mathbf{u}_{1}=-\left[\frac{\mathbf{G m}_{1} \mathbf{m}_{2}}{\mathbf{I}_{12}}+\frac{\mathbf{G}_{\mathbf{m}_{2}} \max }{\mathbf{r}_{22}}+\ldots\right]\)
(vili) If the body of mass \(m\) is moved from a point at a distance \(r_{1}\) to a point at distance \(r_{2}\left(r_{1}>r_{2}\right)\) then change in potential energy
\(\Delta \mathrm{U}=\int_{\mathrm{r}_{1}}^{\mathrm{r}_{2}} \frac{\mathrm{GMm}}{\mathrm{x}^{2}} \mathrm{dx}=-\mathrm{GMm}\left[\frac{1}{\mathrm{r}_{2}}-\frac{1}{\mathrm{r}_{1}}\right]\)
or \(\Delta \mathrm{U}=\mathrm{GMm}\left[\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right]\)
As \(\mathrm{r}_{1}\) is greater than \(\mathrm{r}_{2}\), the change in potential energy of the body will be negative. It means that if a body is brought closer to earth it's potential energy decreases.
(ix) Relation between gravitational potential energy and potential
\(\mathbf{U}=-\frac{\mathbf{G M m}}{\mathbf{r}}=\mathbf{m}\left[\frac{-\mathbf{G M}}{\tau}\right]\)

\(\therefore \mathrm{U}=\mathrm{mV}\)
(x) Gravitational potential energy at the centre of earth relative to infinity.
\(\mathrm{U}_{\text {centre }}=\mathrm{mV}_{\text {centre }}=\mathrm{m}\left(-\frac{3}{2} \frac{\mathrm{GM}}{\mathrm{R}}\right)=-\frac{3}{2} \frac{\mathrm{GMm}}{\mathrm{R}}\)
(xi) Gravitational potential energy of a body at height \(h\) from the earth surface is given by
\(\mathbf{U}_{\mathbf{h}}=-\frac{\mathbf{G} \mathbf{M}_{m}}{\mathbf{R}+\mathbf{h}}=-\frac{\varepsilon \mathbf{R}^{2} \mathbf{m}}{\mathbf{R}+\mathbf{h}}=-\frac{\mathbf{m} \mathbf{g} \mathbf{R}}{1+\frac{\mathbf{b}}{\mathbf{n}}}\)
Escape Velocity The minimum velocity with which a body must be projected up so as to enable it to just overcome the gravitational
pull, is known as escape velocity. The work done to displace a body from the surface of earth ( \(r=R\) ) to infinity \((r=\infty)\) is
\(\mathbf{W}=\int_{\mathbb{R}}^{\infty} \frac{G M m}{x^{2}} d x=-G M m\left[\frac{1}{\infty}-\frac{1}{R}\right]\)
\(\Rightarrow \mathbf{W}=\frac{\mathrm{GMm}}{\mathrm{R}}\)
This work required to project the body so as to escape the gravitational pull is performed on the body by providing an equal amount of kinetic energy to it at the surface of the earth. If ve is the required escape velocity, then kinetic energy which should be given to the body is
\(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{\mathrm{GMm}}{\mathrm{R}} \Rightarrow \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\)
\(\Rightarrow \mathrm{v}_{2}=\sqrt{2 \mathrm{gR}} \quad\left[\mathrm{As} \mathrm{GM}=\mathrm{gR}^{2}\right]\)
or \(\mathbf{v}_{\mathbf{e}}=\sqrt{2 \times \frac{4}{3} \pi \rho \operatorname{GR} \times \mathbf{R}} \Rightarrow \mathbf{v}_{\mathbf{e}}=\mathbf{R} \sqrt{\frac{8}{3} \pi \mathbf{G} \rho} \quad\left[\text { As } \mathbf{g}=\frac{4}{3} \pi \rho \mathbf{G} \mathbf{R}\right]\)
(i) Escape velocity is independent of the mass and direction of projection of the body.
(ii) Escape velocity depends on the reference body. Greater the value of \(\left(\frac{M}{R}\right)\) or \((\mathrm{gR})\) for a planet, greater will be escape velocity.
(iii) For the earth as \(\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\) and \(\mathrm{R}=6400, hence escape velocity of earth is 11.2 m/s

Average Power in LCR circuit
\(\mathbf{P}=\mathbf{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi\)
\(=\left(\frac{100}{\sqrt{2}}\right)\left(\frac{100}{\sqrt{2}}\right) \times 10^{-3} \cos \frac{\pi}{3}\)
\(=2.5 \mathrm{W}\)
KEY CONCEPTS
Power Factor Average power consumed in a cycle \(=V_{r m s} I_{r m s} \cos \phi\)
Here \(\cos \phi\) is called power factor.
Note: I \(\sin \phi\) is called "wattless current"

Average Power in LCR circuit Power consumed of supplied in an AC circuit:
Consider an electrical device which may be a source, a capacitor, a resistor, an inductor or any combination of these. let the potential difference be \(V=V_{A}-V_{B}=V_{m} \sin \sin \omega t .\) let the current through it be \(i=I_{m} \sin \sin (\omega t+\phi)\) instantaneous power \(P\) consumed by the device \(=V_{i}=\left(V_{m} \sin \sin \omega t\right)\left(I_{m} \sin \sin (\omega t+\phi)\right)\)

Average power consumed in a cycle \(=\frac{\frac{2 \pi}{5} \mathrm{Pdt}}{\frac{2 \pi}{\omega}}=\frac{1}{2} V_{m} I_{m} \cos \phi\)
\(=\frac{V_{\mathrm{m}}}{\sqrt{2}} \cdot \frac{I_{\mathrm{m}}}{\sqrt{2}} \cos \phi=V_{r m s} I_{\mathrm{rms}} \cos \phi\)
Here \(\cos \phi\) is called power factor.
Note : \(1 \sin \phi\) is called "wattless current"

Let q be the charge on capacitor plates, then required ratio \(=\frac{\frac{\mathrm{q}}{\mathrm{K}_{1} \varepsilon_{0} \mathrm{A}} \cdot \mathrm{a}}{\frac{\mathrm{q}}{\mathrm{K}_{2} \cdot_{0} \mathrm{A}} \cdot \mathrm{b}}=\frac{\mathrm{K}_{2} \mathrm{a}}{\mathrm{K}_{1} \mathrm{b}}\)
KEY CONCEPTS
Parallel Plate Capacitor Parallel Plate Capacitor It consists of two parallel metallic plates (may be circular, rectangular, square) separated by a small distance

\(A=\) Effective overlaping area of each plate \(\mathrm{d}=\) Separation between the plates \(Q=\) Magnitude of charge on the inner side of each plate \(\sigma=\) Surface density of charge of each plate \(\left(=\frac{Q}{A}\right)\) \(\mathrm{V}=\) Potential difference across the plates \(\mathrm{E}=\) Electric field between the plates \(\left(=\frac{\sigma}{\varepsilon_{0}}\right)\) Capacitance : \(\mathrm{C}=\frac{\varepsilon_{0} A}{d}\) in \(\mathrm{C} . \mathrm{G}\). S. \(\cdot \mathrm{C}=\frac{A}{4 \pi d}\) If a dielectric medium of dielectric constant \(\mathrm{K}\) is filled completely between the plates then capacitance increases by \(\mathrm{K}\) times \(C^{\prime}=K C\)
Dielectric Dielectric

Dielectrics are insulating (non-conducting) materials which transmits electric effect without conducting.
Dielectrics are of two types
(1) Polar dielectrics: A polar molecule has permanent electric dipole moment ( \(\rho\) ) in the absence of electric field also, But a polar dielectric has net dipole moment zero in the absence of electric field because polar molecules are randomiy oriented as shown in figure.

In the presence of electric field polar molecules tends to line up in the direction of electric field,
and the substance has finite dipole moment e.g. water, Alcohol, \(C O_{2}, N H_{3}, H C I\) etc. are made of polar atoms/molecules
(2) Non polar dielectric: In non-polar molecules, each molecule has zero dipole moment in its normal state. When electric field is applied, molecules becomes induced electric dipole e.g. \(\mathrm{N} 2, \mathrm{O} 2,\) Benzene, Methane etc. are made of non-polar atoms molecules
In general, any non-conducting, material can be called as a dielectric but broadly non conducting material having non polar molecules referred to as dielectric.

Applying continuity equation at 1 and \(2,\) we have
\(\mathrm{A}_{1} \mathrm{V}_{1}=\mathrm{A}_{2} \mathrm{V}_{2}\)
(i)
Further applying Bernoulli's equation at these two points, we have
Solving Eqs. (i) and (ii), we have
\(v_{2}^{2}=\frac{2 g h}{1-\frac{A_{2}^{2}}{A_{1}^{2}}}\)
Substituting the values, we have
\(v_{2}^{2}=\frac{2 \times 10 \times 2.475}{1-(0.1)^{2}}=50 m^{2} \cdot s^{-2}\)